https://file.aiquickdraw.com/mathbot/file/1728374336467qdzm3uox.jpg

I see you've uploaded an image. I'll analyze the content of the image to extract and solve the mathematical problem depicted.

Let's begin.The image contains a mathematical equation:

\[
\frac{dy}{dx} = e^{x-y}
\]

This is a separable differential equation, which means we can rearrange it to separate variables \( x \) and \( y \). Here's how to solve it step-by-step:

### Step 1: Rewrite the equation to separate variables
We want all the \( y \)-related terms on one side and the \( x \)-related terms on the other side. Start by rewriting the equation:

\[
\frac{dy}{dx} = e^{x-y} = e^x e^{-y}
\]

Now, separate the variables:

\[
e^y \, dy = e^x \, dx
\]

### Step 2: Integrate both sides
Integrate both sides of the equation:

\[
\int e^y \, dy = \int e^x \, dx
\]

On the left-hand side, the integral of \( e^y \) is \( e^y \), and on the right-hand side, the integral of \( e^x \) is \( e^x \):

\[
e^y = e^x + C
\]

where \( C \) is the constant of integration.

### Step 3: Solve for \( y \)
Now, take the natural logarithm of both sides to solve for \( y \):

\[
y = \ln(e^x + C)
\]

Thus, the solution to the differential equation is:

\[
y = \ln(e^x + C)
\]

This is the general solution to the given differential equation.

Do you need further details or clarification? Here are some additional questions to explore:

1. What happens to the solution when \( C = 0 \)?
2. How can we check the solution by differentiating it?
3. What is the particular solution if an initial condition is given, such as \( y(0) = 1 \)?
4. What would the graph of \( y = \ln(e^x + C) \) look like?
5. How can we interpret this solution in terms of exponential growth or decay?

**Tip:** When solving separable differential equations, always make sure to check whether the variables can be separated before starting the integration process.

Solve the differential equation dy/dx = e^(x-y).

Explore the step-by-step solution to the differential equation dy/dx = e^(x-y) using separation of variables. This solution involves key calculus concepts, including the integration of exponential functions, suitable for college-level students.

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